What is the vertices f(x)=x^2+6x-8 ?

The vertices f(x)=x^2+6x-8 is Minimum (-3,-17)

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vertices f(x)=x^2+6x-8

vertices

f (x) = x^{2} + 6 x - 8

Minimum (- 3, - 17)

Solution steps

y = x^{2} + 6 x - 8

The parabola parameters are:

a = 1, b = 6, c = - 8

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{6}{2 \cdot 1}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 17

Therefore the parabola vertex is

(- 3, - 17)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 17)

f (x) = (x - 3)^{2} - 2

f (x) = x^{2} + 6 x + 4

f (x) = 2 x^{2} - 10 x + 12

\frac{x ^{3} - 1}{x - 1}

+ 10 x^{2} - 15 x + 2

What is the vertices f(x)=x^2+6x-8 ?
The vertices f(x)=x^2+6x-8 is Minimum (-3,-17)